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DochEvi [55]
3 years ago
13

Why is a car racetrack built with a steep inward slope?

Physics
1 answer:
Step2247 [10]3 years ago
5 0
<span>So we want to know why is a car racetrack built with a steep inward slope. When a car is moving on a circular path, it experiences a centrifugal force. This force is bigger as the car speed rises. So to counteract that force the steep inward slope is creating a force component towards the center of the circle that comes from the car's own weight so the car doesn't slide off the road. </span>
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The difference between the Carbon 12 and Carbon 14 isotopes is the
Elina [12.6K]
Answer:
The number of neutrons.

Explaination:
The atoms of carbon 12 contain 6 neutrons while the atoms of carbon 14 contain 8 neutrons
4 0
3 years ago
How long will it take for a car to travel 200 km if it has an average speed of 55 km/hr? ​
andreyandreev [35.5K]

Answer: Approximately 3.65 hours

Explanation:

55 km/h x 3.65 hrs = 200.75 Km/h

4 0
3 years ago
Two electric charges are moved so that they are twice as far apart as they had originally been. Is the force they experience fro
Natasha_Volkova [10]

Answer:

No.

Explanation:

The force that two particle experience is inversely proportional to the sqare of the distance, this is:

F \  \alpha \  \frac{1}{D^{2}} for a distance D

If we move them so that D is doubled:

\frac{1}{2^{2}.D^{2}  }= \frac{1}{4} \eq  \frac{1}{.D^{2}  } \eq

Then the force they experience is one fourth of the original.

5 0
3 years ago
A 2.00-m long uniform beam has a mass of 4.00 kg. The beam rests on a fulcrum that is 1.20 m from its left end. In order for the
Shalnov [3]

Answer:

x ’= 1,735 m,  measured from the far left

Explanation:

For the system to be in equilibrium, the law of rotational equilibrium must be fulfilled.

Let's fix a reference system located at the point of rotation and that the anticlockwise rotations have been positive

             

They tell us that we have a mass (m1) on the left side and another mass (M2) on the right side,

the mass that is at the left end x = 1.2 m measured from the pivot point, the mass of the right side is at a distance x and the weight of the body that is located at the geometric center of the bar

           x_{cm} = 1.2 -1

          x_ {cm} = 0.2 m

          Σ τ = 0

          w₁ 1.2 + mg 0.2 - W₂ x = 0

          x = \frac{m_1 g\ 1.2 \ + m g \ 0.2}{M_2 g}

          x = \frac{m_1 \ 1.2 \ + m \ 0.2 }{M_2}

let's calculate

          x = \frac{2.9 \ 1.2 \ + 4 \ 0.2 }{8.00}2.9 1.2 + 4 0.2 / 8

           

          x = 0.535 m

measured from the pivot point

measured from the far left is

           x’= 1,2 + x

           x'=  1.2 + 0.535

           x ’= 1,735 m

8 0
3 years ago
If 5 mm of rain falls in a 100 m2 field, what volume of rain, in m3, fell in the field?
Nina [5.8K]

The volume of rain that fells in the field is simply given by the area of the field, which is

A=100 mm^2

multiplied by the height of rain that fell, which is

h=5.0 mm

Therefore, the volume is

V=hA=(5 mm)(100 mm^2)=500 mm^3

7 0
3 years ago
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